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Calculating quality factor in a series LCR circuit?

  1. May 29, 2013 #1
    1. The problem statement, all variables and given/known data
    My tutor's notes say that quality factor is supposed to be = 2π (Max. energy stored)/(energy dissipated in one cycle) . So for a standard series LCR circuit, he says:
    max energy = 0.5 L I2
    energy lost per cycle = 0.5 R I2 2π/ω .
    ..and at resonance he calculates it to be equal to be [tex] \frac{1}{R} \sqrt{L/C} [\tex]

    2. Relevant equations
    -

    3. The attempt at a solution
    However, this is how I tried to do it:
    max energy = 0.5 C V2
    energy lost per cycle = 0.5 V2 /R x 2π/ω
    ...and at resonance this gives a quality factor of [tex] R \sqrt{C/L} [\tex]

    Erm.... what da heck?
     
    Last edited: May 29, 2013
  2. jcsd
  3. May 29, 2013 #2
    Latex...y u no work???
     
  4. May 29, 2013 #3
    I'll post the answers here :
    Q from the 1st bit: 1/R x (L/C)^(1/2)
    Q from my try: R x (C/L)^(1/2)
     
  5. May 29, 2013 #4

    rude man

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    .

    Q = wL/R = 1/wRC, w = 1/sqrt(LC)
    Q = 2pi(peak energy stored within one cycle)/(energy dissipated during that cycle.)

    Now, about your computation: You are using the wrong V_max across the capacitor to compute max stored energy on the capacitor.

    max V across capacitor = V0/wRC if the driving voltage is V0 cos(wt).
    What is dissipation in R in terms of V0?

    Take it from there.
     
  6. May 30, 2013 #5
    Max Energy: 0.5 C (V/ωRC)2
    Energy lost per cycle: V2/2R x 2π/ω
    ...and this gives the correct value for Q.

    Thank you.
     
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